Problem 33: Digit cancelling fractions

The fraction frac{49}{98} is a curious fraction, as an inexperienced mathematician in attempting to simplify it
may incorrectly believe that frac{49}{98} = frac{4}{8}, which is correct, is obtained by cancelling the 9s.
We shall consider fractions like, frac{30}{50} = frac{3}{5}, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

My Algorithm

The original problem can be solved with brute force:

two nested loops iterate over all numerators n and denominators d such that n < d

each number is split into its digits

actually only erasing the lower digit of d and the higher digit of n can produce a valid result

multiply all ns and ds, then divide by their Greatest Common Divisor

Hackerrank modified the problem such that larger numbers (up to 4 digits) can be found in both numerator and denominator.
On top of that, there is a variable number of digits to be cancelled.
And the worst: Hackerrank's problem description is very vague and doesn't clarify many corner cases.
Nevertheless, it was way more fun than the simple original problem ...

My main insight was that instead of cancelling/removing digits we can do the inverse, too:
iterate over all "small" numbers and insert digits at all possible positions.

Now we have five (instead of two) nested loops:

the outer loops generate all combinations of n and d such that n < d.

the "middle" loop generates all potential numbers to be inserted; they may have multiple digits

the inner loops produce all permutations of the digits to be inserted

I convert my numbers to std::strings (num2str and str2num).
A string can either be a valid number or contain dots which are placeholders and mean "any digit" - inspired by the syntax of regular expressions.
Note: The placeholder must be alphabetically lower than all digits because I use std::next_permutation:

If we cancel two digits, the middle loop emits "..10", "..11" ... "..99" and the inner loops permute them to
(for "..10":) ".10.", ".1.0", ... "10..".

merge combines a mask (like ".1.0") and a number (like "34") to a number 3140.
until the "large" numerator/denominator match the "small" numerator/denominator.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):

This is equivalent toecho "3 1" | ./33

Output:

(please click 'Go !')

(this interactive test is still under development, computations will be aborted after one second)

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

#include<iostream>

#include<string>

#include<algorithm>

#include<unordered_set>

// convert number to string

std::stringnum2str(unsignedint x, unsignedint digits)

{

std::string result;

// it's faster to generate the digits in reverse order ...

while (digits--> 0)

{

auto digit = x % 10;

result += char(digit +'0');

x /= 10;

}

// ... and then reverse them into their original order

std::reverse(result.begin(), result.end());

return result;

}

// ... and back

unsignedintstr2num(conststd::string& str)

{

unsignedint result = 0;

for (auto s : str)

{

result *= 10;

result += s -'0';

}

return result;

}

// fill all gaps in mask (marked as '.') with the digits found in str and return result as a number

unsignedintmerge(conststd::string& strFill, conststd::string& mask)

{

auto iteFill = strFill.begin();

unsignedint result = 0;

for (auto m : mask)

{

result *= 10;

// if placeholder '.' is found, then take next digit from strFill

if (m =='.')

result += *iteFill++-'0';

else// else take the digit of the mask

result += m -'0';

}

return result;

}

intmain()

{

//#define ORIGINAL

#ifdef ORIGINAL

// brute-force solution for the original problem

unsignedint multD = 1;

unsignedint multN = 1;

for (unsignedint d = 10; d <= 99; d++) // denominator

for (unsignedint n = 10; n < d; n++) // numerator

for (unsignedint cancel = 1; cancel <= 9; cancel++)

{

auto lowN = n % 10;

auto lowD = d % 10;

auto highN = n / 10;

auto highD = d / 10;

// we could check all combinations:

// 1. cancel low digit of denominator and low digit of numerator

// 2. cancel high digit of denominator and low digit of numerator

// 3. cancel low digit of denominator and high digit of numerator

// 4. cancel high digit of denominator and low digit of numerator

// but actually only case 2 is relevant

// (you can prove that but in this problem I focus on the much harder Hackerrank version)

// two fractions a/b and c/d are equal if a*d=b*c

if (highD == cancel && lowN == cancel && lowD * n == highN * d)

{

multN *= n;

multD *= d;

}

}

// shorter code than applying the "least common multiple"

for (unsignedint i = 2; i <= multN; i++)

// remove all common prime factors

while (multN % i == 0 && multD % i == 0)

{

multN /= i;

multD /= i;

}

std::cout<< multD <<std::endl;

return 0;

#endif

// and now a completely different approach for the Hackerrank version of the problem

unsignedint digits;

unsignedint cancel;

std::cin>> digits >> cancel;

auto keep = digits - cancel;

constunsignedint Tens[] = { 1, 10, 100, 1000, 10000 };

unsignedint sumN = 0;

unsignedint sumD = 0;

// don't count fractions twice

std::unordered_set<unsignedint> used;

// I do the inverse:

// instead of removing digits, I add digits

// "d" and "n" stand for denominator and numerator

// they are small numbers where I insert digits

// let's iterate over all "reduced" number and then iterate over all digits we could insert

// note: initially n and d started at Tens[keep - 1] instead of 1 but I learnt the hard way

// that Hackerrank thinks 3016/6032 = 01/02 is a valid reduction

for (unsignedint d = 1; d < Tens[keep]; d++)

for (unsignedint n = 1; n < d; n++)

{

// convert to string

auto strN = num2str(n, keep);

auto strD = num2str(d, keep);

// try to insert all combinations

for (auto insert = Tens[cancel - 1]; insert < Tens[cancel]; insert++)

{

// convert to string

auto strInsert = num2str(insert, cancel);

// if number's digits are (partially) descending, then we already saw all its permutations

bool isAscending = true;

for (size_t i = 1; i < strInsert.size(); i++)

if (strInsert[i - 1] > strInsert[i])

{

isAscending = false;

break;

}

if (!isAscending)

continue;

// prepend placeholders (must be alphabetically smaller than '0')

strInsert.insert(0, keep, '.');

// check all permutations

// strInsertN is permutated until we arrive at the original value again

auto strInsertN = strInsert;

do

{

auto newN = merge(strN, strInsertN);

// the leading digit of the not-cancelled fraction must not be zero

// strangely enough, the leading digit of the cancelled fraction can be zero

if (newN < Tens[digits - 1])

continue;

// strInsertD is permutated until we arrive at the original value again

Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

Changelog

Hackerrank

Difficulty

5%
Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as hard.

Note:Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own.
Maybe not all linked resources produce the correct result and/or exceed time/memory limits.

Heatmap

Please click on a problem's number to open my solution to that problem:

green

solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too

yellow

solutions score less than 100% at Hackerrank (but still solve the original problem easily)

gray

problems are already solved but I haven't published my solution yet

blue

solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much

orange

problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte

red

problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too

black

problems are solved but access to the solution is blocked for a few days until the next problem is published

[new]

the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.

The 310 solved problems (that's level 12) had an average difficulty of 32.6&percnt; at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of &approx;60000 in August 2017)
at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.All of my solutions can be used for any purpose and I am in no way liable for any damages caused.You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.Thanks for all their endless effort !!!

more about me can be found on my homepage,
especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
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